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• Jan 9th 2010, 11:46 PM
Punch
Can't start
Given that $\displaystyle sin(A+B) = 2sin(A-B)$, express $\displaystyle tan A$ in terms of $\displaystyle tan B$
• Jan 9th 2010, 11:59 PM
Sudharaka
Dear Punch,

$\displaystyle Sin(A-B)=SinACosB-SinBCosA$

$\displaystyle Sin(A+B)=SinACosB+SinBCosA$

Given the above relations I think you should be able to express TanA interms of TanB.
• Jan 10th 2010, 12:01 AM
Prove It
Quote:

Originally Posted by Punch
Given that $\displaystyle sin(A+B) = 2sin(A-B)$, express $\displaystyle tan A$ in terms of $\displaystyle tan B$

$\displaystyle \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \mp \cos{\alpha}\sin{\beta}$.

Given that $\displaystyle \sin{(A + B)} = 2\sin{(A - B)}$

$\displaystyle \sin{A}\cos{B} - \cos{A}\sin{B} = 2(\sin{A}\cos{B} + \cos{A}\sin{B})$

$\displaystyle \sin{A}\cos{B} - \cos{A}\sin{B} = 2\sin{A}\cos{B} + 2\cos{A}\sin{B}$

$\displaystyle -3\cos{A}\sin{B} = \sin{A}\cos{B}$

$\displaystyle \frac{-3\sin{B}}{\cos{B}} = \frac{\sin{A}}{\cos{A}}$

$\displaystyle -3\tan{B} = \tan{A}$.
• Jan 10th 2010, 12:04 AM
Punch
Thanks Prove it.. i feel so grateful to you

I have another question,

Given that cos x = p, express sin4x in terms of p
• Jan 10th 2010, 12:09 AM
Sudharaka
Dear Punch,

$\displaystyle Sin4x=2Sin2xCos2x$

$\displaystyle Sin4x=2(2SinxCosx)Cos2x$

$\displaystyle Sin4x=4\sqrt{(1-Cos^2x)}Cosx(2Cos^2x-1)$

By substituting Cosx=p and further simplification will give you the answer.

Hope this helps.
• Jan 10th 2010, 12:14 AM
Punch
Quote:

Originally Posted by Sudharaka
Dear Punch,

$\displaystyle Sin4x=2Sin2xCos2x$

3.$\displaystyle Sin4x=2(2SinxCosx)Cos2x$

4.$\displaystyle Sin4x=4\sqrt{(1-Cos^2x)}Cosx(2Cos^2x-1)$

By substituting Cosx=p and further simplification will give you the answer.

Hope this helps.

I don't get how you get from the 3 to the 4
• Jan 10th 2010, 12:16 AM
Prove It
Quote:

Originally Posted by Punch
Thanks Prove it.. i feel so grateful to you

I have another question,

Given that cos x = p, express sin4x in terms of p

$\displaystyle \sin{4x} = 2\sin{2x}\cos{2x}$

$\displaystyle = 2\cdot 2\sin{x}\cos{x}(\cos^2{x} - \sin^2{x})$

$\displaystyle = 4\cos{x}\sqrt{1 - \cos^2{x}}[\cos^2{x} - (1 - \cos^2{x})]$

$\displaystyle = 4\cos{x}(2\cos^2{x} - 1)\sqrt{1 - \cos^2{x}}$

$\displaystyle = 4p(2p^2 - 1)\sqrt{1 - p^2}$.
• Jan 10th 2010, 12:20 AM
Sudharaka
Dear Punch,

Do you know that,

$\displaystyle Sin2A=2SinACosA$

$\displaystyle SinA=\sqrt{1-Cos^2A}$

These are the two equations I used. In case you do not know them I think you should refer List of trigonometric identities - Wikipedia, the free encyclopedia When going from 3 to 4 I used the second identity.

Hope this helps.
• Jan 10th 2010, 12:33 AM
Punch
got it! thanks!
• Jan 10th 2010, 06:34 AM
Grandad
Hello everyone

If I might make a slight correction:
Quote:

Originally Posted by Prove It
$\displaystyle \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \mp \cos{\alpha}\sin{\beta}$...

should, of course, be:
$\displaystyle \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}$
So the final result will be:
$\displaystyle 3\tan B = \tan A$
Grandad